如图7-223,⊙O1和⊙02外切于点P,直线AD交两圆于A、B、C、D四点,公切线交AD于E点,AB=CD,PE平分∠

1)圆切线性质 PE^2=AE*EB=DE*EC 因为AB=CD=X
所以PE^2=(X+EB)*EB=(X+EC)*EC
X*EB+EB^2=X*EC+EC^2
X*(EB-EC)=EC^2-EB^2
=(EC+EB)(EC-EB)
=-(EC+EB)(BE-EC)
此式有两解
一解得X=-(EC+EB)